the first summand on the right-hand side of the last inequality is smaller than by (1.A.2), the third summand is smaller than by the definition of and, if we refine the gauge , we may suppose, by the definition of , that the second summand is smaller than , and the proof is complete.
We show next that Lemma 1.93 remains valid if we replace by , that is, if instead of the space of Kurzweil–McSchane integrable functions, we consider the space of Henstock–McSchane integrable functions.
Lemma 1.94:Let be a sequence in and be a function. If , then and
Proof. By Lemma 1.93, , and we have the convergence of the integrals. It remains to prove that , that is, for every , there exists a gauge on such that for every -fine ,
However,
Since as tends to infinity, there exists such that the first summand in the last inequality is smaller than for all . Choose an . Then, we can take such that the third summand is smaller than , because it approaches . In addition, once , we can refine so that the second summand becomes smaller than , and we finish the proof.
the first summand on the right-hand side of the last inequality is smaller than 
